Q. The rate of diffusion of methane at a given temperature is twice that of gas $X$. The molecular mass of gas $X$ is
A) 64.0
B) 32.0
C) 4.0
D) 8.0
Explanation
(a)
Let the rate of diffusion of gas $x = a$ and molecular mass $= M$
So, $r_{x} = a,\ M_{x} = M$
${r_{CH}}_{4} = 2a,\ {M_{CH}}_{4} = 16$
$\frac{r_{x}}{{r_{CH}}_{4}} = \sqrt{\frac{{M_{CH}}_{4}}{M_{x}}}$
or $\frac{a}{2a} = \sqrt{\frac{16}{M_{x}}}$ or $M_{x} = 64$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.